CBSE Class 9 Physics Change of State Notes Set 04

States of Matter and Latent Heat

A substance exists as a solid, a liquid or a vapour (gas). To change its state at constant temperature heat must be supplied or removed; this hidden heat is called latent heat. During melting or boiling the temperature stays constant even though heat is being supplied, because the heat goes into changing the state and not into raising the temperature.
1. Fusion (melting) is solid to liquid and freezing is the reverse. Vaporization is liquid to vapour.
2. Heat for a change of state: \( Q = mL \). The SI unit of latent heat is J/kg.
3. Specific latent heat of fusion of ice = 80 cal/g = 336 000 J/kg (80 x 4.2 x 10³). The specific latent heat of vaporization of water is about 2.26 x 10⁶ J/kg (540 cal/g).
4. Latent heat is involved only when the state changes; when a body just becomes hotter or cooler, Q = mc delta T is used.
5. Ice at 0°C cools a substance more effectively than water at 0°C, because it also absorbs the latent heat of fusion.

Worked Examples

Example 1. Heat lost by 5 kg of water at 0°C in becoming ice at 0°C = 5 x 336 000 = 1680 kJ.
Example 2. Heat needed to change 5 kg of ice at -20°C to water at 100°C (c of ice = 2100, c of water = 4200 J/kg K, L = 336 000 J/kg): 5 x 2100 x 20 + 5 x 336 000 + 5 x 4200 x 100 = 210 kJ + 1680 kJ + 2100 kJ = 3990 kJ.
Example 3. 10 g of ice is added to 100 g of water at 30°C. Let the final temperature be t: 100(30 - t) = 10 x 80 + 10t, so t = 20°C.

Calorimetry with Ice

1. Heat lost by the warm body (water and calorimeter) = heat gained by the ice, which first melts (mL) and then warms as water (mc delta T).
2. To find the specific latent heat of ice, add a known mass of ice to warm water in a calorimeter and use this equation with the final temperature.